Question

if (x-a+2b)^2+(x-2a+b)^2=(a+b)^2 the once value of x​

Answer 1

It has given that (x - a + 2b)² + (x - 2a + b)² = (a + b)²

To find : The value of x.

solution : (x - a + 2b)² + (x - 2a + b)² = (a + b)²

using formula, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

so (x - a + 2b)² = x² + a² + 4b² + 2(-xa - 2ab + 2bx)

and (x - 2a + b)² = x² + 4a² + b² + 2(-2ax - 2ab + bx)

⇒x² + a² + 4b² + 2(-xa - 2ab + 2bx) + x² + 4a² + b² + 2(-2ax - 2ab + bx) = a² + b² + 2ab

⇒2x² + 5a² + 5b² - 6ax - 8ab + 6bx = a² + b² + 2ab

⇒2x² + 4a² + 4b² - 6ax - 10ab + 6bx = 0

⇒x² + 2a² + 2b² - 3ax - 5ab + 3bx = 0

⇒x² - (3a - 3b)x + (2a² - 5ab + 2b²) = 0

⇒x² - 3(a - b)x + (2a² - 4ab - ab + 2b²) = 0

⇒x² - 3(a - b)x + (a - 2b)(2a - b) = 0

⇒x² - (a - 2b)x - (2a - b)x + (a - 2b)(2a - b) = 0

⇒[x - (a - 2b)][x - (2a - b)] = 0

⇒x = (a - 2b) , (2a - b)

Therefore the values of x = (a - 2b) , (2a - b)

Answer 2

Answer:

it is the exact answer

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